Explanation Given: The plane z e x 2 − y 2 = 3 Concept Used: Gradient of a function f ( x , y , z ) at a point P ( x 0 , y 0 , z 0 ) is: ∇ f = f x ( x 0 , y 0 , z 0 ) i + f y ( x 0 , y 0 , z 0 ) j + f z ( x 0 , y 0 , z 0 ) k The standard equation of tangent plane to a surface S with equation f ( x , y , z ) = C at point P ( x 0 , y 0 , z 0 ) is f x ( x 0 , y 0 , z 0 ) ( x − x 0 ) + f y ( x 0 , y 0 , z 0 ) ( y − y 0 ) + f z ( x 0 , y 0 , z 0 ) ( z − z 0 ) = 0 Calculation: To find a unit vector, normal to the surface z e x 2 − y 2 = 3 at ( 1 , 1 , 3 ) First, let f ( x , y , z ) = z e x 2 − y 2 Now, find the gradient vector of this function at the point ( 1 , 1 , 3 ) , since gradient vector at any point is perpendicular to the surface at that point. Gradient of a function f ( x , y , z ) is ∇ f = f x ( x 0 , y 0 , z 0 ) i + f y ( x 0 , y 0 , z 0 ) j + f z ( x 0 , y 0 , z 0 ) k Thus, first find f x ( 1 , 1 , 3 ) , f y ( 1 , 1 , 3 ) , and f z ( 1 , 1 , 3 ) as f x ( x , y , z ) = ∂ ∂ x ( z e x 2 − y 2 ) = z ∂ ∂ x e x 2 − y 2 = z e x 2 − y 2 ∂ ∂ x ( x 2 − y 2 ) = z e x 2 − y 2 ( 2 x ) = 2 x z e x 2 − y 2 , f y ( x , y , z ) = ∂ ∂ y ( z e x 2 − y 2 ) = z ∂ ∂ y e x 2 − y 2 = z e x 2 − y 2 ∂ ∂ y ( x 2 − y 2 ) = z e x 2 − y 2 ( − 2 y ) = − 2 y z e x 2 − y 2 and f

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 11.6, Problem 24PS

To determine

To find a unit vector that is normal to the surface

zex2−y2=3 at (1,1,3)

And to find the standard form of the equation of tangent plane to this surface at above point

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Chapter 11.6, Problem 23PS

Chapter 11.6, Problem 25PS

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Chapter 11 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

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To find a unit vector that is normal to the surface z e x 2 − y 2 = 3 at ( 1 , 1 , 3 ) And to find the standard form of the equation of tangent plane to this surface at above point

Solution Summary: The author explains how to find a unit vector that is normal to the given surface and the standard form of the equation of tangent plane to this surface.

Videos

To find a unit vector that is normal to the surface

zex2−y2=3 at (1,1,3)

And to find the standard form of the equation of tangent plane to this surface at above point

Want to see the full answer?

Chapter 11 Solutions

To find a unit vector that is normal to the surface z e x 2 − y 2 = 3 at ( 1 , 1 , 3 ) And to find the standard form of the equation of tangent plane to this surface at above point

Solution Summary: The author explains how to find a unit vector that is normal to the given surface and the standard form of the equation of tangent plane to this surface.

Videos

To find a unit vector that is normal to the surface zex2−y2=3 at (1,1,3) And to find the standard form of the equation of tangent plane to this surface at above point

Want to see the full answer?

Chapter 11 Solutions

To find a unit vector that is normal to the surface

zex2−y2=3 at (1,1,3)

And to find the standard form of the equation of tangent plane to this surface at above point